sallyhansen

not a very helpful forum. If anyone decides they need to get some better explanation please visit this forum http://www.chemicalforums.com/index.php?topic=67705.30#top To get the questions that I have asked in a reasonable amount of time. Thanks for the help I got, but not as useful as I thought it would be! :@

The Ka value is 1.8 x 10-5 do I plugged that into the Ka value part to find the answer for X? So I would assume that I am titrating a weak acid, HA, with NaOH. HA <===> H+ + A- Ka = [H+][A-] / [HA] The term "half titration" simply means that half as much NaOH is added as it would take to reach the end point. You normally would not do this type of titration, but it is useful to determine the experimental value of a Ka. At the half-titration point, the amount of HA left is equal to the amount of A- formed. At this point, the Ka = [H+]. Ka = [H+][A-] / [HA] = [H+] (when A- equals HA) If the pH of the solution is measured at this point, it equals the pKa. If Ka = [H+], the pKa = pH. Thus, a half titration can be used to determine an experimental Ka value. Of course, you have to do a complete titration first, then redo the titration but add only have as much NaOH to reach the half titration point. Okay so for the part of the first question I find confusing is that from question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X. Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1. So the first wuestion I had was this 1. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2) Initial pH = 2.58 pH = -log[H+]=2.58 = [H+] = 10^-2.58 = 2.63*10^-3 m so What I would do is (2.63 x 10^-3)/0.1 = 0.0263 or 2.63 x 10^-2 Would that be the correct answer then? whoops the answer is 6.0 x 10-9

So I am not understanding how I am supposed to approach the question anymore So the answer to the question is correct? Because what I do not get is that I have to find the Ka value of the acetic acid and I plugged in the value of acetic acid to find the x value, so how would that answer the question? I have a few more questions 8. Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1) Original volume of NaOH = 12 Half the volume of NaOH = 6 pH for 12 mL of NaOH = 10.66 pH for 6 mL of NaOH = 4.58 9. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2) So I am not sure about this one would I make the 0.1 mol/L halved? Making it 0.05 mol/L or would I do the question like this 0.1 mol/L of NaOH and 6 mL of NaOH  0.006L (0.006L)(0.1mol/L) =0.0006 moles CH3COOH (aq) + NaOH (aq)  NaCH3COO(aq) + H2O(l) From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0006 moles of acid. n=CV 0.0006 = C(0.006) C= 0.1 mol/L And if that were correct would I not be completing the same ice table twice?

Okay I have finally answered this question (I know I am slow ...) but here is it please tell me if I have completed it correctly 1. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2) HC2H3O2(aq) + H2O(l) à H3O+(aq) + C2H3O2-(aq) HC2H3O2 H3O+ C2H3O2- Initial 0.1 0 0 Change -x +x +x Equilibrium 0.1-x (the x is relatively small compared to 0.1) x x Ka = [H3O+][C2H3O2-] HC2H3O2 Ka= [x][x] [0.1] (1.8 x 10-5)(0.1)= x2 1.8 x 10-6= x2 (square root both sides) x = 1.342 x10-3

okay the concentration is 0.1 mol/L

so once I find the concentration I would use the equation that was on the scan

No but for this question how would I go about answering this 2. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2) How do I calculate Ka? HC2H3O2(aq) + H2O(l) ---> H3O+(aq)+ C2H3O2-(aq) Ka = [H3O+(aq)][ C2H3O2-(aq)]____ [HC2H3O2(aq)]

omg this is excellent. So the part after the pH is the second question onwards correct? Also do you mind continuing you are a life saver. I am not really sure what the amount of ionization will be but I will put the procedure of the lab below. Procedure: 1. Record the molar concentration of the NaOH solution. 0.1 mol/L 2. Produce a table to record your data. It should have one column for volume of NaOH added and one column for pH. 3. Obtain 50 mL of acetic acid and place it into a beaker. 4. Place 50.0 mL of NaOH into the burette. 5. Pipet 25.0 mL of acetic acid into the Erlenmeyer flask. Add two drops of phenolphthalein to the acid. 6. Record the initial pH of the solution. 7. Add 1.00 mL of NaOH from the burette to the Erlenmeyer until the pH reaches 5.00. Record the volume to two decimal places. Measure the pH of the solution each time you add NaOH. 8. Above pH=5.00, add NaOH in 0.10 or 0.20 mL portions. Record the volume at which the phenolphthalein turns pink. phenolphthalein turned pink at 12.0 mL 9. Continue to add NaOH until the pH reaches 11.00. Above pH=11.00, add 0.10 mL portions until the pH reaches 12.00. Note: All you actually have to do is start the burette running, the values will record as you progress through the lab.

Alright so I have been posting these questions up on a whole bunch of forums and no one seems to be answering any of the questions I have posted up, so it would be much appreciated if someone where to help me out with the following questions, because I feel they are easier then they seem but I just over analyze things. Any help on the following questions will be appreciated. Thanks in advance 1. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2) Initial pH = 2.58 2. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2) HC2H3O2(aq) + H2O(l) ---> H3O+(aq)+ C2H3O2-(aq) Ka = [H3O+(aq)][ C2H3O2-(aq)]____ [HC2H3O2(aq)] 3. Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1) half of it would make it 6 mL 4. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2) 5. Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value. (3) 6. Do the values you calculated for [H3O+] and [CH3COOH] prove that CH3COOH is a weak acid? Explain. (2) You see that a majority of the questions are only worth 2 marks, how can that be when there is so much that should be done for them?